Cats and Dogs

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Feb 11, 2010 8:39 am

On monday, a certain animal shelter housed 55 cats and dogs. By friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?

A. 11
B. 12
C. 13
D. 14
E. 20

OA: C

Source: Princeton Review

Please explain!

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Source: Beat The GMAT — Problem Solving | Thu Feb 11, 2010 8:39 am

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Thu Feb 11, 2010 8:57 am

papgust wrote:On monday, a certain animal shelter housed 55 cats and dogs. By friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?

A. 11
B. 12
C. 13
D. 14
E. 20

OA: C

Source: Princeton Review

Please explain!

say C is the no of cats and D is the number of Dogs

C+D = 55

D should be a multiple of 4 (D/4 is an integer) and C should be a multiple of 5 (C/5 is an integer)

We are better with a combination which has D the maximum

D cannot be 52 because C cannot be 3
D cannot be 48 because C cannot be 7
D cannot be 44 because C cannot be 11

D can be 40 since C can be 15

No of pets adopted in this case = 40/4 +15/3 = 13

Always borrow money from a pessimist, he doesn't expect to be paid back.

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lsommer28: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Thu Mar 26, 2009 10:44 am

by lsommer28 » Thu Feb 11, 2010 8:57 am

You have a total of 55 Cats and Dogs. By friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted. That means there has to be at least 51 cats and 4 dogs or 51 dogs and 4 cats. The reason for this is because 1/5 and 1/4 were adopted (this number must be a whole number because you cannot adopt part of an animal). There for to get the greatest number you need to adopt 12 dogs 51/4 = 12 and 1 cat 4/4 = 1.

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Thu Feb 11, 2010 9:05 am

lsommer28 wrote:You have a total of 55 Cats and Dogs. By friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted. That means there has to be at least 51 cats and 4 dogs or 51 dogs and 4 cats. The reason for this is because 1/5 and 1/4 were adopted (this number must be a whole number because you cannot adopt part of an animal). There for to get the greatest number you need to adopt 12 dogs 51/4 = 12 and 1 cat 4/4 = 1.

See the bold part, 51/4 is not a whole number; it is 12.75, since one cant adopt exactly 12.75 dogs; no of dogs cannot be 51

Always borrow money from a pessimist, he doesn't expect to be paid back.

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Thu Feb 11, 2010 10:21 am

papgust wrote:On monday, a certain animal shelter housed 55 cats and dogs. By friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?

A. 11
B. 12
C. 13
D. 14
E. 20

OA: C

Source: Princeton Review

Please explain!

See here C+D=55,
animals adopted = 1/5 C+ 1/4 D. since animals must be in whole no. So C, D are multiple of 5,4 respectively
so C=5,10,15,....55 D=4,8,12....56 so possible combination is (C,D)=(35,20) or (15,40)--> {(1/5)C,(1/4)D} = (7,5)or(3,10).
so max possible animals adopted = 10+3 =13

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amittilak: Senior | Next Rank: 100 Posts; Posts: 34; Joined: Wed Nov 11, 2009 11:23 amFollowed by: 1 member; Thanked: 8 times

by amittilak » Thu Feb 11, 2010 11:25 am

Here's my 2 cents:
C + D = 55
AFTER ADOPTION,
MAXIMIZE, C/5 + D/4
SINCE, 1/5 < 1/4
The answer lies between, C/4 + D/4 AND C/5 + D/5 OR Between 55/4 and 55/5 or between 13.75 and 11.
Now, we know that our answer has to be an integer lower than 13.75 and greater than 11. I chose 13 and did a quick back calculation,
If, C/5 + D/4 = 13,
4C + 5D = 260
4C + 4D = 220
Subtracting, we get D = 40 and C = 15.
Hence, answer is 13 or C

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Thu Feb 11, 2010 11:28 am

lsommer28 wrote:You have a total of 55 Cats and Dogs. By friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted. That means there has to be at least 51 cats and 4 dogs or 51 dogs and 4 cats. The reason for this is because 1/5 and 1/4 were adopted (this number must be a whole number because you cannot adopt part of an animal). There for to get the greatest number you need to adopt 12 dogs 51/4 = 12 and 1 cat 4/4 = 1.

Will you please elaborate your explanation. Though you have reached to ans anyway but nothing is clear what u wanna say.
Also no. of cats and dogs in max. animal adopted is 3 cats and 10 dogs. which does not match with your solution.
So please elaborate your approach and also check for correct solution..

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Feb 11, 2010 8:59 pm

Ajith, nice explanation! Now it seems pretty easy to me than it was previously

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